Prove Cauchy Sequence Convergence: {p_n} \rightarrow p

[mynameisfunk]

Homework Statement

Suppose that {$$p_n$$} is a Cauchy sequence and that there is a subsquence {$$p_{n_i}$$} and a number $$p$$ such that $$p_{n_i} \rightarrow p$$. Show that the full sequence converges, too; that is $$p_n \rightarrow p$$.

Homework Equations

The Attempt at a Solution

Take $$\varepsilon > 0$$. take $$N$$ s.t. $$n_k,n > N$$ implies that $$d(p_{n_k},p)< \frac{\varepsilon}{2}, d(p_n,p_{n_k}) < \frac{\varepsilon}{2}$$. Hence $$d(p_n,p) \leq d(p_{n_k},p)+d(p_{n_k},p_n) \leq \varepsilon$$ Thus {$$p_n$$} converges to $$p$$.

 

[radou]

Actually, $$d(p_n,p) \leq d(p_{n_k},p)+d(p_{n_k},p_n) < \varepsilon$$.

 

[mynameisfunk]

isnt that what i wrote?

 

[mynameisfunk]

Is the only thing i need to fix the scrictly less than inequality??

 

[radou]

Is the only thing i need to fix the scrictly less than inequality??

Yes. Without strictly less, you're not consistent with the definition of convergence.